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# Inequality for experts only...?

Hi, I found this really tough inequality at a blog and nobody has solved it yet (including me)....Can somebody please help me with this please and if possible write a full solution??here ir is:

"Show that (a^b)+(b^a) >1 , for every a,b belonging to (0,1)"

Please give it a try...Thank you very much in advance.....

Could you please try to prove it formally as well......????

Thank you in advance...

a and b do not take the value 0 or 1...

Mo that's the idea of the excercise but can you complete the solution??

### 10 Answers

- Anonymous1 decade agoFavorite Answer
Here is a clean proof using Bernoulli's inequality:

As long as a,b both ∈ (0,1), then there exist some A & B such that A, B are both greater than 1 and a=1/A and b=1/B.

Moreover, since A,B>1, then there exist A1 & B1 both positive, such that A=1+A1 and B=1+B1.

Taking the above notes into account, we conclude that the inequality under consideration turns to

1/(1+A1)^b+1/(1+B1)^a>1.

Let's employ a well known inequality from calculus, the so-called Bernoulli inequality (precisely, a particular form thereof):

For each non-zero x>-1 and c ∈(0,1), we have

(1+x)^c<1+cx.

Therefore (1+A1)^b<1+bA1 and (1+B1)^a<1+aB1 and hence

1/(1+A1)^b>1/(1+bA1) and 1/(1+B1)^a>1/(1+aB1). Adding the latter two together:

1/(1+A1)^b+1/(1+B1)^a>

>1/(1+bA1)+1/(1+aB1).

OK. But A1=1/a-1 and B1=1/b-1. Then

1/(1+bA1)+1/(1+aB1)=a/(a+b-ab)+

+b/(a+b-ab)

=(a+b)/(a+b-ab)

=1+(ab)/(a+b-ab)=1+1/(1/a+1/b-1).

Since a and b ∈ (0,1), then 1/a and 1/b are greater than 1 and therefore 1/a+1/b-1>1 and hence positive. This means that 1+1/(1/a+1/b-1)>1. Q.E.D.

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- 1 decade ago
I'm not an expert on this, but I played with it for fun.

a^b + b^a > 1

ln(a^b + b^a) > ln(1)

ln(a^b) + ln(b^a) > 0

b*ln(a) + a*ln(b) > 0

b*ln(b) > -a*ln(a)

(expr 1): b*ln(b)/a*ln(a) > -1

Note that:

ln(x) = 0 for x = 1

ln(x) > 0 for x > 1

ln(x) < 0 for x >= 0 < 1

ln(x) approaches -infinity when x approaches 0

ln(x) is imaginary for x < 0

So if a and b are > 1 then expr 1 is true

And

if a and b are > 0 < 1 then expr 1 is true.

And

if a and b = 0 then expr 1 is true.

And

if a and b = 1 then expr 1 is true

And

if b = 1 and a > 0 expr 1 is true

And

if a or b < 0 then expr 1 is false

if b > 1 and a >=0 < 1 it gets tricky:

if b*ln(b) >= 1 and -1 < a*ln(a) < 0 then expr 1 will be false

If b >= 1.763 and a < 1 then expr 1 is false

So:

if b > 1.763 and a > 1 then expr 1 is true

It seems as though there are infinite possibilities and several different ranges of values that will work. I guess I'm not sure what "belonging to (0,1)" means exactly.

Update:

If a and b are > 0 < 1 then expr 1 is true because the LHS of expr 1 will always be positive and hence greater than RHS of expr 1 (-1).

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- Scarlet ManukaLv 71 decade ago
Candide: Because most of the answers are wrong.

clavdivs: in fact a^b and b^a will always both be less than 1.

David G: you need to check the interior of the region, not just the boundaries. Also, your approach is not sufficient to check even that the limit of f(a, b) exists at any of the boundary points. You are only checking the limit along one path to each point (three paths at the origin, two paths at the other corners).

M C is checking limits even less rigorously than David G, and also ignores the interior.

I was working on a traditional calculus approach and got as far as proving that the only interior critical point is at (1/e, 1/e), which is a saddle point; I was working on a good proof for the boundaries, but Dale's solution is much better. And it is correct.

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- David GLv 61 decade ago
Strictly speaking a^b + b^a ≥ 1 ∀ a,b ∈ [0, 1]

Let f(a, b) be a^b + b^a

lim x → 0 x^x = 1 by convention (and sound mathematical analysis). It is only at the point a = b = 0 that we have any difficulty, and the limit form removes this. f(0,0) = 2

a^0 = 1 by convention

0^x = 0 by convention (again, there are sound mathematical reasons why this should be)

In the limit ∀ a ≠ 0 as b goes to zero, lim f(a, b) = 1

In the limit as b goes to 1, lim f(a, b) = a + 1

So for all a,b in [0, 1] f(a, b) ≥ 1

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- MattyLv 41 decade ago
If you graph a^x where a <1, you'll notice that it behaves differently from the normal graph of a^x where a>1

Most importantly it always crosses the y axis at 1, so for small x you're returning a value close to one

In the extreme case, as a-->0 the graph of a^x is approaching a straight line, so any value of x will return a value near one, and that value of x raised to the small value of a, because smaller exponent means closer to one, will return another number close to one, the sum being >1

It's pretty intuitive in that sense, but I don't wanna prove it formally

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- MoLv 41 decade ago
This answer will probably be incomplete, but maybe it'll give an idea.

Take the partial wrt a to get

b*a^(b-1) + lnb * b^a = 0

so

a^b = -a/b * lnb * b^a

Substitute into the original:

b^a * (1 - a/b * lnb)

Take partial wrt b:

a * b^(a-1) * (1 - a/b * lnb) + b^a * (a/b^2 * lnb - a/b^2) = 0

b^a * (a/b - a^2/b^2 * lnb +a/b^2 *lnb - a/b^2) = 0

a/b - a^2/b^2 * lnb +a/b^2 *lnb - a/b^2 = 0

b - a lnb + lnb - 1 = 0

a = 1 + (b-1)/lnb

-----------

Tired....

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- 1 decade ago
Why does everyone who answers this question automatically get "thumbs down"?

Warning to other people before they spend time answering it, the author is probably responsible (unless he can give everyone who has answered so far a "thumbs up" without all the "thumbs down"'s being reduced (thus showing he hasn't allocated any thumbs).

Source(s): I wanted to help, but I think its fake interest.- Login to reply the answers

- UnknownDLv 61 decade ago
This is true for all positive numbers.

I think I remember asking a question on this, wait.

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- clavdivsLv 41 decade ago
For a,b in (0,1), then either c=a^b > 1 or d=b^a >1 for all a and b. Hence c+d>1 always.

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- 1 decade ago
fix b and consider the function f(x) = x^b + b^x

f is clearly contineous

limit f(x) when x approaches 0 is equal to 1

limit f(x) when x approaches 1 is equal to 1 + b >1

so f(x) is always greater than 1

and particulary for x = a

this proves it.

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